제출 #1282561

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1282561		poltomo	Jobs (BOI24_jobs)	C++20	100 / 100	226 ms	8592 KiB

Main

#include<iostream>
#include<vector>
#include<queue>

using namespace std;


// find
int find(vector<int>& U, int a) {
	while (a != U[a])
		a = U[a];
	return a;
}
 
// find + two pass path compression
int find2(vector<int>& U, int a) {
	int root = a;
	while (root != U[root])
		root = U[root]; 
	int next = U[a];
	while (next != root) {
		U[a] = root;
		a = next;
		next = U[a];
	}
	return root;
}
 
// union
void merge(vector<int>& U, int a, int b) {
	U[find(U,b)] = find(U, a);
}

int main() {
	int N;
	long long s;
	cin >> N >> s;
	vector<long long> x(N + 1);
	vector<int> p(N + 1);
	vector<long long> cost_to_take(N + 1);
	vector<int> U(N + 1);
	for (int i = 0;i < N + 1;++i) {
		U[i] = i;
	}
	auto cmp = [&](int a, int b) {return cost_to_take[a] > cost_to_take[b];}; // comparator for "least expensive jobs first" behavior by priority queue
	priority_queue<int, std::vector<int>, decltype(cmp)> pq(cmp);
	x[0] = s;
	p[0] = 0;
	for (int i = 1;i < N + 1; ++i) {
		cin >> x[i] >> p[i];
		if (x[i] > 0) {
			pq.push(i); // only jobs with positive profit are worth merging up
		}
		else {
			cost_to_take[i] = -x[i]; // -x(i) is the cost to take a job with non-positive profit, i.e. x(i) <= 0
		}
	}
	while (!pq.empty()) {
		int least_expensive_job = pq.top(); // least expensive job; this job's profit is always positive because only positive profit jobs are pushed onto the priority queue
		pq.pop();
		int least_expensive_job_parent = find2(U, p[least_expensive_job]); // least expensive job's parent job
		if (cost_to_take[least_expensive_job_parent] + x[least_expensive_job_parent] >= cost_to_take[least_expensive_job]) {
		  // Case 1: if this job's parent were to be taken, there would at the very least be cost_to_take[job's parent] + x[job's parent] euros to take children jobs after.
		  // So, if this job costs less than, then it should be taken/merged. Remember, it's profit is positive.
		  // Note, that the very first jobs ever pushed onto the priority queue will cost 0 to take, so they'll all get merged.
			merge(U, least_expensive_job_parent, least_expensive_job); // merge jobs
			x[least_expensive_job_parent] += x[least_expensive_job]; // add profits
			x[least_expensive_job] = 0;
		}
  		else {
  		// Case 2: What does it mean if all case 1 job merges are exhausted? It means that cost_to_take[job's parent] + x[job's parent] < cost_to_take[job] is true for all jobs now.
  		// That means that the next merge will certainly increase the parent job's cost to take. Greedily picking the least expensive positive profit job (case 1 before case 2) brings about the smallest increase in the next merged job's cost to take.
  		// However, tt turns out that greedily always picking the least expensive positive profit job to take is enough, even though it may not fully exhaust the case 1 jobs, in which ... >= cost_to_take[job]).
  		// But it does not matter because there is never a case where taking the least costly to take positive profit job unduly prempts the taking of another or raises its parents cost to take more than it should.
			if (least_expensive_job_parent != 0) {
				cost_to_take[least_expensive_job_parent] = cost_to_take[least_expensive_job] - x[least_expensive_job_parent];
				merge(U, least_expensive_job_parent, least_expensive_job); // merge jobs
				x[least_expensive_job_parent] += x[least_expensive_job]; // add profits
				x[least_expensive_job] = 0;
			}
		}
		if (least_expensive_job_parent != 0 && x[least_expensive_job_parent] > 0) { // the root job has no parent job to merge with and only positive jobs are worth merging
			pq.push(least_expensive_job_parent);
		}
	}
	cout << x[0] - s << endl;
}	

• Subtask 1: 11 / 11
• Subtask 2: 14 / 14
• Subtask 3: 15 / 15
• Subtask 4: 29 / 29
• Subtask 5: 31 / 31